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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 3323 | Accepted: 919 |
Description
Input
Output
Sample Input
9 3 1 1 1 2 2 2 1 1 1 1 2 1 2 3 2 1 2 1 1 1 1 2 2 2 1 1 1
Sample Output
3
SB题，还花了好长时间，不开心，不写题解了。
#include <iostream> #include <cstring> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> #include <time.h> #include <string> #include <map> #include <stack> #include <vector> #include <set> #include <queue> #define inf 0x3f3f3f3f #define mod 10000 typedef long long ll; using namespace std; const int N=505; const int M=250005; int s,t,n,m,cnt,tim,top,cut,k; int head[M],dfn[M],low[M],stack1[M]; int num[M],in[M],out[M],vis[M],w[N][N]; int dis[4][2]= {0,1,1,0,-1,0,0,-1}; bool flag=false; struct man { int to,nxt; } edg[M*10]; void addedg(int u,int v) { edg[cnt].to=v; edg[cnt].nxt=head[u]; head[u]=cnt++;//printf("!!!%d %d\n",u,v);system("pause"); } void init() { cnt=0; tim=0; top=cut=k=0; memset(head,-1,sizeof head); memset(dfn,0,sizeof dfn); memset(low,0,sizeof low); memset(stack1,0,sizeof stack1); memset(num,0,sizeof num); memset(in,0,sizeof in); memset(out,0,sizeof out); memset(vis,0,sizeof vis); memset(edg,0,sizeof edg); memset(w,0,sizeof w); } void Tarjan(int u) { int v; low[u] = dfn[u] = ++tim; stack1[top++] = u; vis[u] = 1; for(int e = head[u]; e != -1; e = edg[e].nxt) { v = edg[e].to; if(!dfn[v]) { Tarjan(v); low[u] = min(low[u], low[v]); } else if(vis[v]) { low[u] = min(low[u], dfn[v]); } } if(low[u] == dfn[u]) { cut++; do { v = stack1[--top]; num[v] = cut; vis[v] = 0; }while(u != v); } } void build(int i,int j,int d) { int xx=i+dis[d][0]; int yy=j+dis[d][1]; int u=i*m+j,v=xx*m+yy; if(xx>=0&&yy<m&&yy>=0&&xx<n){ if(w[i][j]>=w[xx][yy])addedg(u,v); if(w[i][j]<=w[xx][yy])addedg(v,u); } return; } int main() { while(~scanf("%d%d",&m,&n)) { init(); for(int i=0; i<n; i++) { for(int j=0; j<m; j++) { scanf("%d",&w[i][j]); } } for(int i = 0; i < n; i++) { for(int j = 0; j < m; j++) { for(int d = 0; d < 2; d++) { build(i,j,d); } } } for(int i=0; i<n*m; i++)if(!dfn[i])Tarjan(i); for(int i=0; i<n*m; i++) { for(int j=head[i]; j!=-1; j=edg[j].nxt) { int v=edg[j].to; if(num[i]!=num[v])out[num[i]]++,in[num[v]]++; } } int father=0,son=0; for(int i=1; i<=cut; i++) { if(in[i]==0)father++; if(out[i]==0)son++; } if(cut==1)printf("0\n"); else printf("%d\n",max(father,son)); } return 0; }
POJ2375 Cow Ski Area （强连通）（缩点）
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原文地址：http://www.cnblogs.com/jianrenfang/p/5893731.html
